18.100B, FALL 2002
SOLUTIONS TO RUDIN, CHAPTER 4, PROBLEMS 2,3,4,6
Problem 2
If f : X −→ Y is a continuous map then f −1 (C) ⊂ X is closed for each closed
subset C ⊂ Y. For any map and any subset G ⊂ Y , f (f −1 (G)) = G. Now, if E ⊂ X
then C = f (E) is closed and E ⊂ f −1 (C) (since x ∈ E implies f (e) ∈ f (E) ⊂ C
implies e ∈ f −1 (C)). By the continuity condition f −1 (C) is closed so E ⊂ f −1 (C)
which implies f (E) ⊂ f (E).
Consider X = [0, 1) ∪ [1, 2] as a subset of R with the usual metric. Then f :
X −→ [0, 1] given by
(
x x ∈ [0, 1)
f (x) =
1 x ∈ [1, 2]
is continuous since a convergent sequence {xn } in X is eventually in [0, 1) or [1, 2]
and so {f (xn )} converges by the continuity of x and constant maps. On the other
hand, E = [0, 1) is closed and f (E) is not, with f (E) = [0, 1] so f (E) is strictly
contained in f (E).
Problem 3
By definition Z(f ) = f −1 ({0}). The set {0} is closed and f is continuous, so
Z(f ) is closed.
Problem 4
If y ∈ f (X) then there exists x ∈ X such that f (x) = y. By the density of E
in X there is a sequence {xn } in E with xn → x in X. By the continuity of f,
f (xn ) → f (x) = y so f (E) is dense in f (X).
Suppose g(p) = f (p) for all p ∈ E. Given x ∈ X, by the result above, there exists
{xn } in E such that xn → x and f (xn ) → f (x). The continuity of g means that
g(xn ) = f (xn ) → g(x) so f (x) = g(x) for all x ∈ X.
Problem 6
The distance on X × Y is the sum of the distances on X and Y. I will do it with
sequences.
Suppose E is compact and f : E −→ Y is continuous. Now suppose {pn } is a
sequence in graph(f ). Thus, pn = (xn , f (xn )) for some sequence {xn } in E. By the
compactness of E, there is a convergent subsequence {xn(k) }. By the continuity of
f, f (xn(k) ) is convergent, and hence pn = (xn(k) , f (xn(k) )) is convergent, so each
sequence in graph(f ) has a convergent subsequence. It follows that it is compact.
Conversely, suppose that E and graph(f ) are both compact. Let {xn } be a
convergent sequence in E, xn → x. Then {(xn , f (xn ))} is a sequence in graph(f )
so by its compactness has a convergent subsequence, (xn(k) , f (xn(k) )) → (x, q).
Since the graph is closed, this must be a point in it, so q = f (x). This argument
1
2
18.100B, FALL 2002 SOLUTIONS TO RUDIN, CHAPTER 4, PROBLEMS 2,3,4,6
applies to any subsequence of {xn }, so we see that any subsequence of {f (xn )} has
a convergent subsequence with limit f (x). This however implies that f (xn ) → f (x),
since if not there would exist a sequence f (xn(k) with d(f (x), f (xn(k) ) ≥ c > 0 and
this cannot have such a convergent subsequence. Thus in fact xn → x implies that
f (xn ) → f (x), so f is continuous.